Question

Kinda tricky integral:
\[\int_{-\infty}^\infty \frac{\cos x}{(x^2+1)(1+e^x)}\,dx\]

Answer 1

I've got some hints in mind, but I'm debating whether or not they give too much away :)

Answer 2

*

Answer 3

I don't know but I kind of feel like it has something to do with Gaussian integral.

Answer 4

It's related in that the result also contains \(\pi\), but beyond that I'm not sure... It'd be fascinating to see a derivation of this integral using the Guassian.

Answer 5

Convert to polar coordinate analogous to how you would evaluate Gaussian integral?

Answer 6

\[
f(x)=\frac{\cos x}{(x^2+1)(1+e^x)}\\
\begin{align*}
f(x)f(y)&=\frac{\cos x}{(x^2+1)(1+e^x)}\frac{\cos y}{(y^2+1)(1+e^y)}\\
&=\frac{\cos(x)\cos(y)}{\left(r^2+x^2y^2+1)(1+e^x+e^y+e^{x+y}\right)},\,r^2=x^2+y^2
\end{align*}
\]
Aesthetically pleasing, probably of no use.

Answer 7

Alright, here's my first hint: Notice that \(\dfrac{\cos x}{x^2+1}\) is even.

Answer 8

Hint (1.a): \(y=-x\)

Answer 9

Picking up on that hint :

\[I=\int_{-\infty}^\infty \frac{\cos x}{(x^2+1)(1+e^x)}\,dx\]

sub \(y=-x \) and get
\[I=\int_{-\infty}^\infty \frac{e^x \cos x}{(x^2+1)(1+e^x)}\,dx\]

Add them and get
\[2I =\int_{-\infty}^\infty \frac{\cos x}{x^2+1}\,dx \]

Answer 10

since the integrand is even,
\[I =\int_{0}^\infty \frac{\cos x}{x^2+1}\,dx \]

Answer 11

please do not give further hints, I think i got this :)

Answer 12

So far so good! I had a lot of trouble making use of the "hidden" fact that
\[\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}=1\]
The rest of the solution can go any of several ways, I think. I'd love to see what you come up with.

Answer 13

we may try DUT
\[F(y) = \int_{0}^\infty \frac{\cos(xy)}{x^2+1}\,dx\]
\[\implies F''(y) = -\int_{0}^\infty \frac{x^2\cos(xy)}{x^2+1}\,dx \]
\[\implies -F''(y) + F(y) = \int_{0}^\infty \cos(xy)\,dx \]

Answer 14

scratch that, doesn't converge

Answer 15

Oooh, a related problem:
\[\int_0^{\pi/2} \cos(\tan x)\,dx\](which is equivalent to the integral at hand, which you can see if you set \(t=\tan x\))

Answer 16

@ganeshie8 I'm not at all surprised MSE has this one covered, but let's not spoil it for the rest. Based off of one of the answers I skimmed over, you seem to be on the right track with that ODE approach.

Answer 17

I think its mostly done :

\[F''(y) - F(y) = 0 \implies F(y) = C_1e^y + C_2e^{-y}\]

finding \(C_1\) and \(C_2\) by choosing appropriate initial conditions should be a piece of cake

Answer 18

do you not have to subtract them as the integration order needs to be reversed with the sub?

Answer 19

differential also changes sign
so that fixes the bounds..

Answer 20

duh
:-)

Answer 21

The answer is \(\dfrac{\pi}{e}\)? Read something extremely similar to this on Wikipedia and want to confirm the answer before trying.

Answer 22

Did I missed by a factor of 2?

Answer 23

Yep, the value of the integral is \(\dfrac{\pi}{2e}\), but it's not the answer that's important, it's how to get it :)

Answer 24

Wikipedia: Residue theorem.

I don't know how Wikipedia did this but the answer is basically there. I don't even know line integral let alone the complex version of it.

Answer 25

Read at your own discretion!

Answer 26

Right, contour integration is always an option, you just have to choose the right contour. My approach utilized the Laplace transform. First you parameterize the integral as follows.
\[I(a)=\int_{-\infty}^\infty \frac{\cos ax}{(x^2+1)(1+e^x)}\,dx\]
The change of variables/exponential identity reduces this to
\[I(a)=\int_0^\infty \frac{\cos ax}{x^2+1}\,dx\]
Take the Laplace transform:
\[\mathcal{L}_s\{ I(a)\}=\int_0^\infty \left(\int_0^\infty \frac{\cos ax}{x^2+1}\,dx\right)e^{-as}\,da\]
and some rearranging gives
\[\mathcal{L}_s\{ I(a)\}=\int_0^\infty \left(\int_0^\infty e^{-as}\cos ax\,da\right)\frac{dx}{x^2+1}\]
where the inner integral is just the Laplace transform of \(\cos ax\), i.e. \(\mathcal{L}_s\{\cos ax\}=\dfrac{s}{s^2+x^2}\).
\[\mathcal{L}_s\{ I(a)\}=\int_0^\infty \frac{dx}{(x^2+1)(x^2+s^2)}\]
Partial fractions:
\[\frac{1}{(x^2+1)(x^2+s^2)}=\frac{1}{s^2-1}\left(\frac{1}{x^2+1}-\frac{1}{x^2+s^2}\right)\]
And now we can integrate:
\[\begin{align*}\mathcal{L}_s\{ I(a)\}&=\frac{1}{s^2-1}\int_0^\infty \left(\frac{1}{x^2+1}-\frac{1}{x^2+s^2}\right)\,dx\\[1ex]
&=\frac{1}{s^2-1}\left(\frac{\pi}{2}-\frac{\pi}{2s}\right)\\[1ex]
&=\frac{\pi}{2}\left(\frac{1}{s^2-1}-\frac{1}{s(s^2-1)}\right)\\[1ex]
&=\frac{\pi}{2}\left(\frac{1}{s^2-1}+\frac{1}{s}-\frac{s}{s^2-1}\right)
\end{align*}\]
And finally, take the inverse transform.
\[\begin{align*}\mathcal{L}_s\{ I(a)\}&=\frac{\pi}{2}\left(\frac{1}{s^2-1}+\frac{1}{s}-\frac{s}{s^2-1}\right)\\[1ex]
I(a)&=\mathcal{L}_a^{-1}\{\cdots\}\\[1ex]
&=\frac{\pi}{2}\left(\sinh a+1-\cosh a \right)
\end{align*}\]
Let \(a=1\) and we're done.

Answer 27

dark magic